pmo 15 national written

8/9/2019 PMO 15 National Written

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15th Philippine Mathematical Olympiad

National Stage, Written Phase

26 January 2013

Time Allotment: 4 hours Each item is worth 8 points.

1. Determine, with proof, the least positive integer n for which there existn distinct positive integers x1, x2, x3, . . . , xn such that

1 − 1x1

1 − 1

x2

1 − 1

x3

· · ·

1 − 1xn

=

15

2013.

2. Let P be a point in the interior of ABC . Extend AP , BP , and CP tomeet B C , AC , and AB at D, E , and F , respectively. If AP F , BP D,and CP E have equal areas, prove that P is the centroid of ABC .

3. Let n be a positive integer. The numbers 1, 2, 3, . . . , 2n are randomlyassigned to 2n distinct points on a circle. To each chord joining twoof these points, a value is assigned equal to the absolute value of the

difference between the assigned numbers at its endpoints.Show that one can choose n pairwise non-intersecting chords such thatthe sum of the values assigned to them is n2.

4. Let a, p, and q be positive integers with p ≤ q . Prove that if one of thenumbers a p and aq is divisible by p, then the other number must also bedivisible by p.

5. Let r and s be positive real numbers that satisfy the equation

(r + s − rs)(r + s + rs) = rs.Find the minimum values of r + s − rs and r + s + rs.


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Problem 1. Determine, with proof, the least positive integer n for whichthere exist n distinct positive integers x1, x2, x3, . . . , xn such that

1 − 1

x11 −

1

x21 −

1

x3 · · ·

1

− 1

xn =

15

2013

.

Solution. Suppose x1, x2, x3, . . . , xn are distinct positive integers that satisfythe given equation. Without loss of generality, we assume that x1 < x2 <x3


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Problem 2. Let P be a point in the interior of ABC . Extend AP , BP ,and CP to meet BC , AC , and AB at D, E , and F , respectively. If AP F ,BP D, and CP E have equal areas, prove that P is the centroid of ABC .Solution. Denote by (XY Z ) the area of

XY Z . Let w = (AP F ) =

(BP D) = (CP E ), x = (BP F ), y = (CP D), and z = (AP E ).Having the same altitude, we get

BD

DC =

(BAD)

(CAD) =

2w + x

w + y + z

andBD

DC =

(BP D)

(CP D) =

w

y,

which implies

wy + xy = w2 + wz. (1)

Similarly, we also get

wz + yz = w2 + wx and wx + xz = w2 + wy. (2)

Combining equations (1) and (2) gives

xy + yz + xz = 3w2

. (3)

On the other hand, by Ceva’s Theorem, we have

AF

F B · BD

DC · CE

EA =

(AP F )

(BP F ) · (BP D)

(CP D) · (CP E )

(AP E ) =

w

x · w

y · w

z = 1, (4)

orw3 = xyz. (5)

Applying equation (5) to equation (3) gives

w

z +

w

x +

w

y = 3. (6)

Equations (4) and (6) assert that the geometric mean and the arithmeticmean of the positive numbers w

x, w

y, and w

z are equal. By the equality condition

of the AM-GM Inequality, it follows that

w

x =

w

y =

w

z = 1 or w = x = y = z.

Therefore, we conclude that AF = F B, BD = DC , and CE = EA, whichmeans that P is the centroid of ABC . q.e.d.


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Problem 3. Let n be a positive integer. The numbers 1, 2, 3, . . . , 2n arerandomly assigned to 2n distinct points on a circle. To each chord joiningtwo of these points, a value is assigned equal to the absolute value of thedifference between the assigned numbers at its endpoints.

Show that one can choose n pairwise non-intersecting chords such that thesum of the values assigned to them is n2.

Solution. First, observe that

ni=1

i = n(n + 1)

2 and

2ni=n+1

i = n2 + n(n + 1)

2 ,

which means that2n

i=n+1

i −n

i=1

i = n2.

Let A = {1, 2, . . . , n} and B = {n + 1, n + 2, . . . , 2n}. (Here, we do notdistinguish the point labeled x and the number x itself.) Because the numbersare arranged on a circle, one can find a pair {x1, y1}, where x1 ∈ A andy1 ∈ B, such that one arc joining x1 and y1 contains no other labeled points.One can then remove the chord (including x1 and y1) joining these points.Among the remaining labeled points, one can find again a pair

{x2, y2

}, where

x2 ∈ A \ {x1} and y2 ∈ B \ {y1}, such that one arc joining x2 and y2 doesnot contain a labeled point, and then remove again the chord (including theendpoints) joining x2 and y2. Continuing this process, one can find pairs{x3, y3}, {x4, y4}, and so on, and then remove the chords joining the pairs.

We claim that the removed chords satisfy the required properties. Clearly,there are n such chords. Because no labeled point lies on one arc joining x jand y j for any 1 ≤ j ≤ n, the removed chords are non-intersecting. Finally,the sum of the values assigned to the removed chords is

n j=1

(y j − x j) =n

j=1

y j −n

j=1

x j =2n

i=n+1

i −n

i=1

i = n2.

This ends the proof of our claim. q.e.d.


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Problem 4. Let a, p, and q be positive integers with p ≤ q . Prove that if one of the numbers a p and aq is divisible by p, then the other number mustalso be divisible by p.

Solution. Suppose that p |

a p. Since p ≤

q , it follows that a p

| aq , which

implies that p | aq .Now, suppose that p | aq , and, on the contrary, p a p. Then there is a

prime number r and a positive integer n such that rn | p (which implies thatrn ≤ p) and rn a p. Since p | aq , it follows that r | a, and so rn | an. Thismeans that p < n, which gives the following contradiction:

2 p ≤ r p < rn ≤ p.

Therefore, a p

must also be divisible by p

. q.e.d.


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Problem 5. Let r and s be positive real numbers that satisfy the equation

(r + s − rs)(r + s + rs) = rs.Find the minimum values of r + s

−rs and r + s + rs.

Solution. The given equation can be rewritten into

(r + s)2 = rs(rs + 1). (1)

Since (r + s)2 ≥ 4rs for any r, s ∈ R, it follows that rs ≥ 3 for any r, s > 0.Using this inequality, equation (1), and the assumption that r and s arepositive, we have

r + s − rs = rs(rs + 1) − rs = 1 1 + 1

rs + 1

≥ 1 1 + 1

3 + 1

= −3 + 2√

3.

Similarly, we also have

r + s + rs ≥ 3 + 2√

3.

We show that these lower bounds can actually be attained. Observe thatif r = s =

√ 3, then

r + s − rs = −3 + 2√

3 and r + s + rs = 3 + 2√

3.

Therefore, the required minimum values of r + s − rs and r + s + rs are−3 + 2√ 3 and 3 + 2√ 3, respectively. q.e.d.

pmo 15 national written

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